When using stokes theorem to remove integrals

1. Apr 4, 2014

joshmccraney

hey pf!

i had a question. namely, in the continuity equation we see that $$\frac{\partial}{\partial t}\iiint_V \rho dV = -\iint_{S} \rho \vec{v} \cdot d\vec{S}$$ and we may use the divergence theorem to have: $$\frac{\partial}{\partial t}\iiint_V \rho dV = -\iiint_{V} \nabla \cdot \big( \rho \vec{v} \big) dV$$
ultimately, we arrive at: $$\frac{\partial}{\partial t}\bigg( \rho \bigg) = -\nabla \cdot \big( \rho \vec{v} \big)$$
my question is, at this point, how are we able to do the following two things:
1 interchange $\frac{\partial}{\partial t}$ inside the volume integral?
2 drop the volume integrals entirely?

i should say that the volume is arbitrary, and from what i remember, we have to do something like $\lim_{V \to 0}$ but i don't know the formal, mathematical procedure here.

thanks!

2. Apr 4, 2014

cwilkins

Here is some typical reasoning:
1. The spatial coordinates themselves are independent of time, so performing time differentiation before or after has no effect on the result (provided the integration is definite).
2. The integrals are over the same *arbitrary* volume. If they are equal then the integrands must indeed be equal. This follows naturally from the fundamental theorem of calculus for definite integration.

Last edited: Apr 4, 2014
3. Apr 4, 2014

joshmccraney

but why do we need to take a limit?